Question

Correct increasing order for the wavelengths of absorption in the visible region for the complexes of $C{o}^{3+}$ is:

• A. $\left[Co\right(H_{2}O{)}_6{]}^{3+},\left[Co\right(en{)}_3{]}^{3+},\left[Co\right(N{H}_3{)}_6{]}^{3+}$
• B. $\left[Co\right(H_{2}O{)}_6{]}^{3+},\left[Co\right(N{H}_3{)}_6{]}^{3+},\left[Co\right(en{)}_3{]}^{3+}$
• C. $\left[Co\right(N{H}_3{)}_6{]}^{3+},\left[Co\right(en{)}_3{]}^{3+},\left[Co\right(H_{2}O{)}_6{]}^{3+}$
• D. $\left[Co\right(en{)}_3{]}^{3+},\left[Co\right(N{H}_3{)}_6{]}^{3+},\left[Co\right(H_{2}O{)}_6{]}^{3+}$

Answer 1

The correct option is D $\left[Co\right(en{)}_3{]}^{3+},\left[Co\right(N{H}_3{)}_6{]}^{3+},\left[Co\right(H_{2}O{)}_6{]}^{3+}$

According to the electrochemical series, ligand strength increases in the order,
$H_{2}O<N{H}_3<en$
Acoordingly increasing order of crystal field splitting energy is:
$H_{2}O<N{H}_3<en$
Again we know,
As, $E=\frac{hc}{\lambda }$
$\text{Where}\lambda \text{is wavelength}$
Thus it is inversely proportional to energy.
Thus, increasing order of wavelength of absorption is:
$\left[Co\right(en{)}_3{]}^{3+}<\left[Co\right(N{H}_3{)}_6{]}^{3+}<\left[Co\right(H_{2}O{)}_6{]}^{3+}$